\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 20, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2012/20\hfil Nonexistence of solutions to BVP]
{Nonexistence of solutions to some boundary-value problems for
second-order ordinary differential equations}
\author[G. L. Karakostas \hfil EJDE-2012/20\hfilneg]
{George L. Karakostas}
\address{George L. Karakostas \newline
Department of Mathematics, University of Ioannina,
451 10 Ioannina, Greece}
\email{gkarako@uoi.gr}
\thanks{Submitted December 9, 2011. Published February 2, 2012.}
\subjclass[2000]{34B15, 34B99, 34K10}
\keywords{Boundary value problems; non-existence of solutions}
\begin{abstract}
We present a method to obtain concrete sufficient conditions
which guarantee non-existence of solutions lying into a prescribed
domain of six two- or three-point boundary value problems for
second-order ordinary differential equations.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks
\section{Introduction}
Let $I$ be the interval [0,1] of the real line $\mathbb{R}$ and
let $C(I)$ be the Banach space of all continuous functions
$x: I\to\mathbb{R}$, endowed with the sup-norm $\|\cdot\|$.
In this article, we investigate the non-existence of a solution $x$
of the equation
\begin{equation}\label{e7}
x''(t)+(Fx)(t)=0,\quad \text{a. a. }t\in{I},
\end{equation}
satisfying one of the following conditions:
\begin{gather}
\label{e1} x(0)=0,\quad x'(1)=\alpha x'(0),\\
\label{e2} x(0)=0,\quad x(1)=\alpha x'(0),\\
\label{e3} x'(0)=0,\quad x(1)=\alpha x(\eta),\\
\label{e4} x'(0)=0,\quad x'(1)=\alpha x(0),\\
\label{e5} x(0)=0,\quad x(1)=\alpha x(\eta), \\
\label{e6} x(0)=\alpha x'(0),\quad x(1)=0,
\end{gather}
and lying into a prescribed domain of the space $C(I)$.
Here assume that $\alpha\geq 0$ and $\eta\in[0,1]$.
The dependence of the response $F$ on the function $x$ might be in a moment
or a functional way.
Some publications which deal with the existence of positive solutions of
equations of the form \eqref{e7} lying in a certain domain, associated
with the conditions (or the multi-point version of them) respectively, are,
for instance, \cite{bf, ma, kt1, kt2, kt3} for \eqref{e1};
\cite{LG1} for \eqref{e2}, \cite{aoy1, JGZ} for \eqref{e3};
\cite{cz1} for \eqref{e4};
\cite{bf1, chen, chen1, CR1, DG1, db1, HG1, KT1, LP1, MW, YR, SW1,
xx, YS, w1, WL1, wi1, ZS1} for \eqref{e5};
\cite{mt} for \eqref{e6}. (See, also, the references therein)
In most of these cases the response factor $(Fx)(t)$ has the
Hammerstein form $q(t)f(x(t))$ and the quotient ${f(u)}/{u}$ plays
a central role. In particular, its least and upper limiting values
at $0+$ and at $+\infty$ are combined with some arguments related
to the Green's function, in order to ensure the applicability of
a method leading to the existence of a fixed point of an appropriate
integral operator. But, as the sufficient conditions are
important for the existence of solutions, the necessary conditions
are (more or less) equally important.
Working in this direction, in this paper
we provide some rather simple sufficient conditions for the nonexistence
of (positive) solutions, which lie in an angular domain of the origin.
Our discussion refers to the existence of a real number $\rho>0$
such that no solution of the problem exists satisfying the inequality
$$
|(Fx)(t)|1$, then there is no positive solution $x$ of
the problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and
such that $(Fx)(t)\geq 0$, a.e. on $I$.
\item[(ii)] If $\alpha\in[0,1]$, then there is no solution $x$ of
problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and such
that
\begin{equation}\label{e01}
\operatorname{ess\,sup}\frac{(Fx)(t)}{x^2(t)}1$ and let $x$ be a positive solution of the
problem in ${\mathcal{D}}(F)$. Consider a real number
$\lambda\neq 0$ and write equation \eqref{e7} in the form
$$
x''(t)+\lambda x'(t)=\lambda x'(t)-(Fx)(t).
$$
Multiply both sides with $e^{\lambda t}$ and take
$$
\big(x'(t)e^{\lambda t}\big)'=[x''(t)+\lambda x'(t)]e^{\lambda t}
=[\lambda x'(t)-(Fx)(t)]e^{\lambda t}.
$$
Integrating from 0 to $t$, we obtain
\begin{align*}
x'(t)e^{\lambda t}&=x'(0)+{\lambda}\int_0^tx'(s)e^{\lambda s}ds
-\int_0^t(Fx)(s)e^{\lambda s}ds.\\
&=x'(0)+\lambda x(t)e^{\lambda t}-\lambda x(0)
-\int_0^tz(s)e^{\lambda s}ds,
\end{align*}
where
$z(s):=\lambda^2x(s)+(Fx)(s)$.
Thus we have
\begin{equation}\label{e8}
x'(t)-\lambda x(t)=x'(0)e^{-\lambda t}
-\int_0^tz(s)e^{-\lambda (t-s)}ds.
\end{equation}
Multiplying both sides by $e^{-\lambda t}$, we obtain
\[
\big(x(t)e^{-\lambda t}\big)'
=[x'(t)-\lambda x(t)]e^{-\lambda t}
=x'(0)e^{-2\lambda t}-\int_0^tz(s)e^{-\lambda (2t-s)}ds.
\]
Integrate both sides from 0 to $t$ and take
\begin{align*}
x(t)e^{-\lambda t}
&=x'(0)\int_0^te^{-2\lambda s}ds-\int_0^t\int_0^rz(s)e^{-\lambda (2r-s)}dsdr\\
&=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t})
-\int_0^tz(s)e^{\lambda s}\int_s^te^{-2\lambda r}drds\\
&=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t})
-\frac{1}{2\lambda}\int_0^tz(s)e^{\lambda s}(e^{-2\lambda s}-e^{-2\lambda t}),
\end{align*}
because of \eqref{e1}.
Finally we obtain
\begin{equation}\label{e9}
x(t)=\frac{x'(0)}{\lambda}\sinh({\lambda t})
-\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.
\end{equation}
By using \eqref{e1} and \eqref{e8} it follows that
\begin{align*}
\alpha x'(0)=x'(1)
&=\lambda x(1)+x'(0)e^{-\lambda}-\int_0^1z(s)e^{-\lambda (1-s)}ds\\
&=x'(0)\cosh({\lambda})-\int_0^1z(s)\cosh({\lambda(1-s)})ds
\end{align*}
and therefore the solution $x$ must satisfy
\begin{equation}\label{e10}
x'(0)(\cosh(\lambda)-\alpha)=\int_0^1z(s)\cosh({\lambda(1-s)})ds,
\end{equation}
for all $\lambda\neq 0$.
Observe that the right side is a positive quantity, while the left
side depends on $\lambda$. Hence, if it holds $x'(0)>0$, choose
$\lambda$ such that $\cosh(\lambda)\alpha$, to get a contradiction.
\smallskip
(ii) Next assume that $\alpha\in[0,1]$. Let $x$ be a solution satisfying
relation \eqref{e01}. Choose $\lambda$ negative large enough such that
$$
z(t):=(Fx)(t)+\lambda x^2(t)<0,
$$
for a.a. $t\in I$. This and \eqref{e10} imply that
$x'(0)<0$ and so, due to the fact that $x(0)=0$, the solution $x$
can not be positive. The proof is complete.
\end{proof}
In the sequel we shall assume that $00$ we shall denote by ${\mathcal{A}}_{\rho}$ the set
of all functions $x\in{\mathcal{D}}(F)$ satisfying the inequality
$$
\|Fx\|_{\infty}0$ for all $\lambda\in[0,1]$.
Also, since $\phi(0)=1$ we conclude that for each $\alpha\in(0,1)$
there is a $\lambda\in (0,1)$ such that
$\alpha\alpha \}
$$
is nonempty and it contains a right neighborhood of 0.
\begin{theorem} \label{thm3.1}
Assume that $F$ satisfies condition {\rm (C)}.
Then there is none $x\in {\mathcal{A}}_{\rho_2}$ which solves
problem \eqref{e2}-\eqref{e7}, where
$$
\rho_2:=2(1-\alpha).
$$
\end{theorem}
\begin{proof}
Assume that a solution exists satisfying the requirements of the theorem
and take $\rho'$ such that $\rho'0$ such that
\begin{equation}\label{e13}
\rho'\lambda\alpha$.
Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$.
Dividing both sides of \eqref{e14} by $x(t_0)$, we obtain
\[
\lambda[\sinh(\lambda)-\lambda\alpha]
\leq\sinh(\lambda)\int_0^1\Big[\lambda^2\frac{x(s)}{\|x\|}
+\frac{|(Fx)(s)|}{\|x\|}\Big]\sinh({\lambda(1-s)})ds.
\]
From this relation we obtain
$$
\lambda(\sinh(\lambda)-\lambda\alpha)
< \sinh({\lambda})\int_0^1[\lambda^2+\rho']\sinh({\lambda(1-s)})ds
$$
and so
$$
\lambda^2(\sinh(\lambda)-\lambda\alpha)
< \sinh({\lambda})[\lambda^2+\rho'](\cosh({\lambda)-1)}.
$$
The latter contradicts to \eqref{e13} and so there is no solution
of the problem.
\end{proof}
\section {Problem \eqref{e3}-\eqref{e7}}
\begin{theorem}\label{th2}
Assume that $F$ satisfies condition {\rm (C)}.
Then there is no $x\in {\mathcal{A}}_{\rho_3}$ that solves
problem \eqref{e22}-\eqref{e1}, where
$$
\rho_3:=\sup_{\lambda>0}\frac{2-\cosh(\lambda)-\alpha
e^{\lambda(\eta-1)}}{\cosh(\lambda)-1} \lambda^2.
$$
\end{theorem}
\begin{proof}
Assume that a solution exists satisfying the requirements of the
theorem and take $\rho'$ such that $\rho'0$ such that
\begin{equation}\label{e15}
\rho'0}\frac{\lambda[e^{\lambda}-\alpha]}{\lambda\cosh(\lambda)
-\lambda+1-e^{-\lambda}}-\lambda^2.
$$
\end{theorem}
\begin{proof}
Assume that a solution $x$ exists satisfying the requirements of the theorem and
take $\rho'$ such that $\rho'0$ such that
\begin{equation}\label{e19}
\rho'0}\frac{\lambda^2[\sinh(\lambda)
-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda)\big(\cosh(\lambda)-1\big)}-\lambda^2.
$$
\end{theorem}
\begin{proof}
Assume that a solution $x$ exists satisfying the requirements of the theorem and
take $\rho'$ such that $\rho'0$ such that
\begin{equation}\label{e21}
\rho'0}\frac{\lambda^2[\alpha\lambda
+\cosh(\lambda)]}{\sinh(\lambda)\big(\lambda\alpha e^{\lambda}
+\cosh(\lambda)\big)}-\lambda^2.
$$
\end{theorem}
\begin{proof}
Assume that a solution $x$ exists satisfying the requirements of the theorem
and take $\rho'$ such that $\rho'0$ such that
\begin{equation}\label{e23}
\rho'